Derivatives — MATHS STD 11 Science — Question

Let $\text{u}=2^\text{x};\text{v}=\cot\text{x};\text{w}=\text{x}^{\frac{-1}{2}}$

Then, $\text{u}'=2^\text{x}\log2;\text{v}=-\text{cosec}^2\text{x};\text{w}'=\frac{-1}{2}\text{x}^{\frac{-3}{2}}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uvw})=\text{u}'\text{vw}+\text{uw}\text{v}'+\text{uv}\text{w}'$

$\frac{\text{d}}{\text{dx}}\Big[2^\text{x}\cot\text{x}\Big(\text{x}^{\frac{-1}{2}}\Big)\Big]=2^{\text{x}}\log2.\cot\text{x}.\text{x}^{\frac{-1}{2}}+2^\text{x}(\text{cosec})^2\text{x}\text{x}^\frac{-1}{2}+2^\text{x}\cot\text{x}\Big(\frac{-1}{2}\text{x}^\frac{-3}{2}\Big)$

$=2^{\text{x}}\log2.\cot\text{x}.{\frac{1}{\sqrt{\text{x}}}}+2^\text{x}(\text{cosec}^2\text{x})\frac{1}{\sqrt{\text{x}}}+2^\text{x}\cot\text{x}\Big(\frac{-1}{2\text{x}\sqrt{\text{x}}}\Big)$

$=\frac{2^\text{x}}{\sqrt{\text{x}}}\Big(\log2.\cot\text{x}-\text{cosec}^2\text{x}-\frac{\cot\text{x}}{2\text{x}}\Big)$